[Ninja_Bait]

Solving each wheel generally for static equilibrium (from quiz 6):

1. Fny=Tau/r
2. Distance from wheel to CoG = sqrt(trackwidth^2+wheelbase^2)/2 and from now on is d
3. Fny*d*sin(angle included) = Fnx*d*sin(other angle included) because net torque is zero.
4. Fny(trackwidth/d) = Fnx(wheelbase/d)
5. Fnx = Fny*f = Tau*f2/r
6. |Fn| = |Fnx + Fny| = Tau*sqrt(f2^2 + 1)/r = Nn*mu

I think we only have to exceed the smallest static friction force (F1 and F2) to create a net torque on the bot so:

7. Tau*sqrt(f2^2 + 1)/r > N1*mu
8. Tau > N1*mu*r/sqrt(f2^2 + 1)

We can solve the forces in the yz plane for one side of the robot:

8. let wheelbase = b and distance from CoM to CoG = x
9. f1 = 2x/b
10. Tau about CoM (in the yz plane) is zero, therefore N1(b/2+x) = N4(b/2-x)
11. Divide by b/2, so N1(1+f1) = N4(1-f1)
12. Fnet is also zero, so W/2 = N1+N4
13. N1 = (W/2-N1)(1-f1)/(1+f1)
14. N1 = (W/4)(1-f1)

Now we can solve Tau in terms of givens:

15. Tau > (W/4)*(1-f1)*mu*r/sqrt(f2^2 + 1)

But I think I'm making the wrong assumption in the second section. If we have to exceed the greater static friction, Tau > (W/4)*(1+f1)*mu*r/sqrt(f2^2 + 1). I'm not so sure how the free-body diagram works out. On the other hand, I could be totally wrong altogether... (Thanks for the solution, ether!)